Respuesta :
Answer:
b) 0.025
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 100, \sigma = 15[/tex]
The proportion of adults with scores above 130 is closest to
This proportion is 1 subtracted by the pvalue of Z when X = 130. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{130 - 100}{15}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a pvalue of 0.9772
1 - 0.9772 = 0.0228
Closest to 0.025
So the correct answer is:
b) 0.025
Answer: option b is the closet
Step-by-step explanation:
Since the scores on the Wechsler Adult Intelligence Scale are approximately Normal, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = scores on the Wechsler Adult Intelligence Scale.
µ = mean score
σ = standard deviation
From the information given,
µ = 100
σ = 15
We want to find the proportion of adults with scores above 130. It is expressed as
P(x > 130) = 1 - P(x ≤ 130)
For x = 130,
z = (130 - 100)/15 = 2
Looking at the normal distribution table, the probability corresponding to the z score is 0.9773
P(x > 130) = 1 - 0.9773 = 0.023
